Question

If the lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}and\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$ intersect, then the value of k is ?

Answer 1

The co-ordinates of any point on the first line are given by

$\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}=\lambda$

$\Rightarrow x-1=2\lambda ,y+1=3\lambda ,z-1=4\lambda$

$\Rightarrow x=1+2\lambda ,y=-1+3\lambda ,z=1+4\lambda$

Thus, the co-ordiantes of any point on this line are $(1+2\lambda ,-1+3\lambda ,1+4\lambda )$

The co-ordiantes of any point on the second line are

$\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}=\mu$

$\Rightarrow x-3=\mu ,y-k=2\mu ,z=\mu$

$\Rightarrow x=3+\mu ,y=k+2\mu ,z=\mu$

Thus, the co-ordiantes of any point on second line are $(3+\mu ,k+2\mu ,\mu )$

If these two lines intersect each other, then

$3+\mu =1+2\lambda$ ,$k+2\mu =-1+3\lambda$ and $\mu =1+4\lambda$

$\Rightarrow 2\lambda -\mu =2$, $3\lambda -2\mu =k+1$ and $4\lambda -\mu =-1$

Solving the equations $2\lambda -\mu =2$ and $4\lambda -\mu =-1$

$\Rightarrow 2\lambda -\mu -4\lambda +\mu =2-(-1)$

$\Rightarrow -2\lambda =2+1=3$ or $\lambda =\frac{-3}{2}$

Substituting the value of $\lambda =\frac{-3}{2}$ in $2\lambda -\mu =2$

we get $2\times \frac{-3}{2}-\mu =2$ or $-3-\mu =2$ or $\mu =-5$

Substituting the values of $\lambda =\frac{-3}{2}$ and $\mu =-5$ in

$3\lambda -2\mu =k+1$

we get $3\times \frac{-3}{2}-2\times -5=k+1$ or $k+1=\frac{-9}{2}+10=\frac{-9+20}{2}=\frac{11}{2}$

$\therefore k+1=\frac{11}{2}$

$\Rightarrow k=\frac{11}{2}-1=\frac{11-2}{2}=\frac{9}{2}$